Simplify and expand the following expression: $ \dfrac{3}{n + 1}+ \dfrac{2}{n - 10}- \dfrac{n}{n^2 - 9n - 10} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{n}{n^2 - 9n - 10} = \dfrac{n}{(n + 1)(n - 10)}$ Now we have: $ \dfrac{3}{n + 1}+ \dfrac{2}{n - 10}- \dfrac{n}{(n + 1)(n - 10)} $ The least common multiple of the denominators is: $ (n + 1)(n - 10)$ In order to get the first term over $(n + 1)(n - 10)$ , multiply by $\dfrac{n - 10}{n - 10}$ $ \dfrac{3}{n + 1} \times \dfrac{n - 10}{n - 10} = \dfrac{3(n - 10)}{(n + 1)(n - 10)} $ In order to get the second term over $(n + 1)(n - 10)$ , multiply by $\dfrac{n + 1}{n + 1}$ $ \dfrac{2}{n - 10} \times \dfrac{n + 1}{n + 1} = \dfrac{2(n + 1)}{(n + 1)(n - 10)} $ Now we have: $ \dfrac{3(n - 10)}{(n + 1)(n - 10)} + \dfrac{2(n + 1)}{(n + 1)(n - 10)} - \dfrac{n}{(n + 1)(n - 10)} $ $ = \dfrac{ 3(n - 10) + 2(n + 1) - n} {(n + 1)(n - 10)} $ Expand: $ = \dfrac{3n - 30 + 2n + 2 - n}{n^2 - 9n - 10} $ $ = \dfrac{4n - 28}{n^2 - 9n - 10}$